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InterviewSolution
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Specific volume of cylindrical virus particle is `6.02xx10^(-2) c c//g` whose radius and length `7 Å` and `10 Å` respectively. If `N_(A)=6.02xx10^(23)`, find molecular weight of virus:A. 1.54 kg/mol.B. `1.54 xx 10^4` kg/mol.C. `3.08 xx 10^4` kg/mol.D. `3.08 xx 10^3` kg/mol. |
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Answer» Correct Answer - A Sp. Vol (vol. of 1gm ) cylindrical virus particle `=6.02xx10^(-2)` cc/gm radius of virus r=7Å= `7xx10^(-8)` cm length of virus =`pir^2l` `=22/7xx(7xx10^(-8))^2xx10xx10^(-8)=154xx10^(-23)` cc wt. of one virus particle =`"Vol."/"Sp. vol."` `rArr (154xx10^(-23))/(6.02xx10^(-2))` gm `therefore` mol. wt. of virus =wt. of `N_A` particles `=(154xx10^(-23))/(6.02xx10^(-2))xx6.02xx10^(+23)` gm/mol =15400 gm/mol =15.4 kg/mol |
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